[Leetcode 19] 19 Remove Nth Node From End Of List

Problem:

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

Analysis:

Use an auxilary pointer that n procedeed the current node. When the aux pointer is null, then the current pointer is the pointer to be removed;

Also need a pointer that 1 succeed the current point for update.

This is a one pass algorithm, so the time complecity is O(n); and space complexity is O(1);

Code:

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
// Start typing your Java solution below
// DO NOT write main() function
ListNode toRmv = head, aux = head, preRmv = head;

for (int i=0; i<n; ++i)
aux = aux.next;

while (aux != null) {
preRmv = toRmv;
toRmv = toRmv.next;
aux = aux.next;
}

if (toRmv == head) {
head = head.next;
} else {
preRmv.next = toRmv.next;
}

return head;
}
}

View Code

Attention:

If the Node to be removed is the head, need process it separately.